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Someone gives you two dice. They offer to pay you €37 if you roll two 6s. However, it will cost you €1 a roll. What is the EV of this wager?

Double sixes = $1/36 = 0.028$

Other rolls = $35/36 = 0.972$

Do we calculate expectation value like this

0.028(€36) + 0.972(-€1) = x

or like this

0.028(€37) + 0.972(-€1) = x

I think we multiply postive outcomes(0.028) by money we will get €37 and substract bad outcomes (0.972) multiplied by €1 that we are loosing. But some say first equation is correct(€36). Are they right and why?

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I think I figure it...if they said they will pay me 37 if I win and I pay them 1 if I loose the my theory would be right –  Shark4Ever Apr 22 '12 at 0:51
    
Correct. The second describes precisely such a situation. –  Cameron Buie Apr 22 '12 at 0:52

1 Answer 1

up vote 1 down vote accepted

You will pay whether you win or lose. The first is correct.

Think of it this way: 1/36 of the time you "should" win, so you "should" win 1/36 of the reward in a given role (what expected value means). Then subtract what you pay. It ends up being

$x=\frac{1}{36}(37)-1=\frac{1}{36}=\frac{1}{36}(36)+\frac{35}{36}(-1)$.

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Yes I figurite out thank you –  Shark4Ever Apr 22 '12 at 0:52

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